I am a newbie in R and searched in several forums but didn't got an answer so far. We are asked to do a maximum likelihood estimation in R for an AR(1) model without using the arima() command. We should estimate the intercept alpha, the coefficient beta and the variance sigma2. The data should be following a normal distribution, where I derived the log-likelihood function from. I was then trying to program the function with the following code:
Y <- data$V2
nlogL <- function(theta,Y){
alpha <- theta[1]
rho <- theta[2]
sigma2 <- theta[3]
logl <- -(100/2)*log(2*pi) - (100/2)*log(theta[3]) - (0.5*theta[3])*sum(Y-(theta[1]/(1-theta[2]))**2)
return(-logl)
}
par0 <- c(0.1,0.1,0.1)
opt <- optim(par0, nlogL, hessian = TRUE)
When running this code I always get the error message: Error in Y - (theta[1]/(1 - theta[2]))^2 : 'Y' is missing.
It would be great if you could have a look whether the likelihood function is derived correctly.
Thank you very much in advance for your help!
Your nlogL function should only take a single argument, theta. So you can fix your immediate problem simply by removing the 2nd argument to the function, and the Y variable would be resolved by its definition outside of nlogL. Alternatively, you could keep the signature of nlogL as-is and pass Y as an additional argument through optim like this: optim(par0, nlogL, hessian = TRUE, Y=Y). Also I would second chinsoon12's suggestion to review ?optim.
Related
I want to find a package in R to fit the extreme value distribution
https://en.wikipedia.org/wiki/Generalized_extreme_value_distribution with three unknown parameters mu, sigma, xi.
I found two packages that can do the inference for these three parameters based on maximum likelihood estimation.
library(ismev)
gev.fit(data)
and
library(extRemes)
fevd(data)
the output is estimates of mu, sigma, and xi.
But if I just want to fit distribution with two parameters mu and sigma (like Gumbel distribution, the parameter xi=0). How to apply the above two packages? Or are there any other packages that can do inference for the Gumbel distribution?
The evd package has 2-parameter [dpqr]gumbel functions that you can combine with any general-purpose optimization method (optim() is one such possibility, as suggested in the comments, but there are some shortcuts as suggested below).
Load packages, simulate example:
library(evd)
library(fitdistrplus)
set.seed(101)
x <- rgumbel(1000, loc = 2, scale = 2.5)
Make a more robust wrapper for dgumbel() that won't throw an error if we hand it a non-positive scale value (there are other ways to deal with this problem, but this one works):
dg <- function(x, loc, scale, log) {
r <- try(dgumbel(x, loc, scale, log), silent = TRUE)
if (inherits(r, "try-error")) return(NA)
return(r)
}
fitdistr(x, dg, start = list(loc = 1, scale = 1))
Results seem reasonable:
loc scale
2.09220866 2.48122956
(0.08261121) (0.06102183)
If you want more flexibility I would recommend the bbmle package (for possibly obvious reasons :-) )
I want to fit a non-linear model to a real data.
The real data consists of 2 known numerical vectors ; thickness as 'x' and fh as 'y'
thickness=seq(0.15,2.00,by=0.05)
fh = c(5.17641, 4.20461, 3.31091, 2.60899, 2.23541, 1.97771, 1.88141, 1.62821, 1.50138, 1.51075, 1.40850, 1.26222, 1.09432, 1.13202, 1.12918, 1.10355, 1.11867, 1.09740,1.08324, 1.05687, 1.19422, 1.22984, 1.34516, 1.19713,1.25398 ,1.29885, 1.33658, 1.31166, 1.40332, 1.39550,1.37855, 1.41491, 1.59549, 1.56027, 1.63925, 1.72440, 1.74192, 1.82049)
plot(thickness,fh)
This is apparently non-linear. So, I am trying to fit this model as a non-linear function of
y= x*2/3+(2+2*a)/(3*x)
Variable a is an unknown constant and I am trying to find the best constant a that minimizes the sum of square of error between the regression line and the real data.
I first used a function fitModel that I found on a YouTube video, Fitting Functions to Data in R.
library(TIMP)
f=fitModel(fh~thickness^2/3+(2+2*A)/(3*thickness)) #it finds the coefficient 'A'
coef(f) # to represent just the coefficient
However, there's an error
Error in modelspec[[datasetind[i]]] : subscript out of bounds
So, as an alternative, want to find a plot of 'a' and 'the Sum of Squares of Error'. This time, I have such a hard time finding 'a' and plotting this graph. By manual work, I figured out the value 'a' is somewhere near 0.2 but this is not a precise value.
It would be helpful if someone could manifest either:
Why the fitModel function didn't work or
How to find the value a and plot the graph.
You could try this instead:
yf = function(a,xv) xv*(2/3)+(2+2*a)/(3*xv)
yf(2,thickness)
f <- function (a,y, xv) sum((y - yf(a,xv))^2)
f(2,fh,thickness)
xmin <- optimize(f, c(0, 10), tol = 0.0001, y=fh,xv=thickness)
xmin
plot(thickness,fh)
lines(thickness,yf(xmin$minimum,thickness),col=3)
I'm trying to find the MLE of distribution whose pdf is specified as 'mixture' in the code. I've provided the code below that gives an error of
"Error in optim(start, f, method = method, hessian = TRUE, ...) :
L-BFGS-B needs finite values of 'fn'"
"claims" is the dataset im using. I tried the same code with just the first two values of "claims" and encountered the same problem, so for a reproducible example the first two values are 1536.77007 and 1946.92409.
The limits on the parameters of the distribution is that 0<.p.<1 and a>0 and b>0, hence the lower and upper bounds in the MLE function. Any help is much appreciated.
#create mixture of two exponential distribution
mixture<-function(x,p,a,b){
d<-p*a*exp(-a*x)+(1-p)*b*exp(-b*x)
d
}
#find MLE of mixture distribution
LL <- function(p,a,b) {
X = mixture(claims,p,a,b)
#
-sum(log(X))
}
mle(LL, start = list(p=0.5,a=1/100,b=1/100),method = "L-BFGS-B", lower=c(0,0,0), upper=c(1,Inf,Inf))
edit: Not really sure why dput(), but anyway,
#first two values of claims put into dput() (the actual values are above)
dput(claims[1:2])
c(307522.103, 195633.5205)
I'm trying to use fitdist () function from the fitdistrplus package to fit my data to different distributions. Let's say that my data looks like:
x = c (1.300000, 1.220000, 1.160000, 1.300000, 1.380000, 1.240000,
1.150000, 1.180000, 1.350000, 1.290000, 1.150000, 1.240000,
1.150000, 1.120000, 1.260000, 1.120000, 1.460000, 1.310000,
1.270000, 1.260000, 1.270000, 1.180000, 1.290000, 1.120000,
1.310000, 1.120000, 1.220000, 1.160000, 1.460000, 1.410000,
1.250000, 1.200000, 1.180000, 1.830000, 1.670000, 1.130000,
1.150000, 1.170000, 1.190000, 1.380000, 1.160000, 1.120000,
1.280000, 1.180000, 1.170000, 1.410000, 1.550000, 1.170000,
1.298701, 1.123595, 1.098901, 1.123595, 1.110000, 1.420000,
1.360000, 1.290000, 1.230000, 1.270000, 1.190000, 1.180000,
1.298701, 1.136364, 1.098901, 1.123595, 1.316900, 1.281800,
1.239400, 1.216989, 1.785077, 1.250800, 1.370000)
Next, if i run fitdist (x, "gamma") everything is fine, but if I use fitdist (x, "beta") instead I get the following error:
Error in start.arg.default(data10, distr = distname) :
values must be in [0-1] to fit a beta distribution
Ok, so I'm not native english but as far as I understand this method requires data to be in the range [0,1], so I scale it by using x_scaled = (x-min(x))/max(x). This gives me a vector with values in that range that perfectly correlates the original vector x.
Because of x_scaled is of class matrix, I convert into a numeric vector using as.numeric(). And then fit the model with fitdist(x_scale,"beta").
This time I get the following error:
Error in fitdist(x_scale, "beta") :
the function mle failed to estimate the parameters, with the error code 100
So after that I've been doing some search engine queries but I don't find anything useful. Does anybody ave an idea of whats going on wrong here? Thank you
By reading into the source code, it can be found that the default estimation method of fitdist is mle, which will call mledist from the same package, which will construct a negative log-likelihood for the distribution you have chosen and use optim or constrOptim to numerically minimize it. If there is anything wrong with the numerical optimization process, you get the error message you've got.
It seems like the error occurs because when x_scaled contains 0 or 1, there will be some problem in calculating the negative log-likelihood for beta distribution, so the numerical optimization method will simply broke. One dirty trick is to let x_scaled <- (x - min(x) + 0.001) / (max(x) - min(x) + 0.002), so there is no 0 nor 1 in x_scaled, and fitdist will work.
I'm trying to use the nls.lm function in the minpack.lm to fit a non-linear model to some data from a psychophysics experiment.
I've had a search around and can't find a lot of information about the package so have essentially copied the format of the example given on the nls.lm help page. Unfortunately my script is still failing to run and R is throwing out this error:
Error in fn(par, ...) :
unused argument (observed = c(0.1429, 0.2857, 0.375, 0.3846, 0.4667, 0.6154))
It appears that the script thinks the data I want to fit the model to is irrelevant, which is definitely wrong.
I'm expecting it to fit the model and produce a value of 0.5403 for the spare parameter (w).
Any help is greatly appreciated.
I'm making the transfer from Matlab over to R so apologies if my code looks sloppy.
Here's the script.
install.packages("pracma")
require(pracma)
install.packages("minpack.lm")
require(minpack.lm)
# Residual function, uses parameter w (e.g. .23) to predict accuracy error at a given ratio [e.g. 2:1]
residFun=function(w,n) .5 * erfc( abs(n[,1]-n[,2])/ ((sqrt(2)*w) * sqrt( (n[,1]^2) + (n[,2]^2) ) ) )
# example for residFun
# calculates an error rate of 2.59%
a=matrix(c(2,1),1,byrow=TRUE)
residFun(.23,a)
# Initial guess for parameter to be fitted (w)
parStart=list(w=0.2)
# Recorded accuracies in matrix, 1- gives errors to input into residFun
# i.e. the y-values I want to fit the model
Acc=1-(matrix(c(0.8571,0.7143,0.6250,0.6154,0.5333,0.3846),ncol=6))
# Ratios (converted to proportions) used in testing
# i.e. the points along the x-axis to fit the above data to
Ratios=matrix(c(0.3,0.7,0.4,0.6,0.42,0.58,0.45,0.55,0.47,0.53,0.49,0.51),nrow=6,byrow=TRUE)
# non-linear model fitting, attempting to calculate the value of w using the Levenberg-Marquardt nonlinear least-squares algorithm
output=nls.lm(par=parStart,fn=residFun,observed=Acc,n=Ratios)
# Error message shown after running
# Error in fn(par, ...) :
# unused argument (observed = c(0.1429, 0.2857, 0.375, 0.3846, 0.4667, 0.6154))
The error means you passed a function an argument that it did not expect. ?nls.lm has no argument observed, so it is passed to the function passed to fn, in your case, residFun. However, residFun doesn't expect this argument either, hence the error. You need to redefine this function like this :
# Residual function, uses parameter w (e.g. .23) to predict accuracy error at a given ratio [e.g. 2:1]
residFun=function(par,observed, n) {
w <- par$w
r <- observed - (.5 * erfc( abs(n[,1]-n[,2])/ ((sqrt(2)*w) * sqrt( (n[,1]^2) + (n[,2]^2) ) ) ))
return(r)
}
It gives the following result :
> output = nls.lm(par=parStart,fn=residFun,observed=Acc,n=Ratios)
> output
Nonlinear regression via the Levenberg-Marquardt algorithm
parameter estimates: 0.540285874836135
residual sum-of-squares: 0.02166
reason terminated: Relative error in the sum of squares is at most `ftol'.
Why that happened :
It seems that you were inspired by this example in he documentation :
## residual function
residFun <- function(p, observed, xx) observed - getPred(p,xx)
## starting values for parameters
parStart <- list(a=3,b=-.001, c=1)
## perform fit
nls.out <- nls.lm(par=parStart, fn = residFun, observed = simDNoisy,
xx = x, control = nls.lm.control(nprint=1))
Note that observed is an argument of residFun here.